Speed Control of DC Motors: the speed of a motor is given by the

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    Speed Control of DC Motors: the speed of a motor is given by the


    aaaa RIVK



    60 where Ra= armature circuit resistance. It is

    obvious that the speed can be controlled by varying

    (i) Flux/pole, (Flux Control)

    (ii) Resistance Ra of armature circuit (Rheostatic Control) and

    (iii) Applied voltage V (Voltage Control).

    Speed Control of Shunt motor:

    (i) Variation of Flux or Flux Control Method: By decreasing the flux, the

    speed can be increased and vice versa. The flux of a dc motor can be

    changed by changing Ish with help of a shunt field rheostat. Since Ish is

    relatively small, shunt field rheostat has to carry only a small current,

    which means I2shR loss is small, so that rheostat is small in size.

    (ii) Armature or Rheostatic Control Method: This method is used when

    speeds below the no-load speed are required. As the supply voltage is

    normally constant, the voltage across the armature is varied by inserting a

    variable rheostat in series with the armature circuit. As controller

    resistance is increased, voltage across the armature is decreased, thereby

    decreasing the armature speed. For a load constant torque, speed is

    approximately proportional to the voltage across the armature. From the

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    speed/armature current characteristic, it is seen that greater the resistance

    in the armature circuit, greater is the fall in the speed.

    (iii) Voltage Control Method:

    (a) Multiple Voltage Control: In this method, the shunt field of the motor

    is connected permanently to a fixed exciting voltage, but the armature is

    supplied with different voltages by connecting it across one of the several

    different voltages by means of suitable switchgear. The armature speed

    will be approximately proportional to these different voltages. The

    intermediate speeds can be obtained by adjusting the shunt field regulator.

    (b) Ward-Leonard System: This system is used where an unusually wide

    and very sensitive speed control is required as for colliery winders,

    electric excavators, elevators and the main drives in steel mills and

    blooming and paper mills. M1 is the main motor whose speed control is

    required. The field of this motor is permanently connected across the dc

    supply lines. By applying a variable voltage across its armature, any

    desired speed can be obtained. This variable voltage is supplied by a

    motor-generator set which consists of either a dc or an ac motor M2

    directly coupled to generator G. The motor M2 runs at an approximately

    constant speed. The output voltage of G is directly fed to the main motor

    M1. The voltage of the generator can be varied from zero up to its

    maximum value by means of its field regulator. By reversing the direction

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    of the field current of G by means of the reversing switch RS, generated

    voltage can be reversed and hence the direction of rotation of M1. It

    should be remembered that motor generator set always runs in the same


    Speed Control of Series Motors:

    1. Flux Control Method: Variations in the flux of a series motor can be

    brought about in any one of the following ways:

    (a) Field Diverters: The series winding are shunted by a variable

    resistance known as field diverter. Any desired amount of current can be

    passed through the diverter by adjusting its resistance. Hence the flux can

    be decreased and consequently, the speed of the motor increased.

    (b) Armature Diverter: A diverter across the armature can be used for

    giving speeds lower than the normal speed. For a given constant load

    torque, if Ia is reduced due to armature diverter, the must increase

    (Ta aI ) This results in an increase in current taken from the supply

    (which increases the flux and a fall in speed (N I/ )). The variation in

    speed can be controlled by varying the diverter resistance.

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    (c) Trapped Field Control Field: This method is often used in electric

    traction. The number of series filed turns in the circuit can be changed.

    With full field, the motor runs at its minimum speed which can be raised

    in steps by cutting out some of the series turns.

    (d) Paralleling Field coils: this method used for fan motors, several

    speeds can be obtained by regrouping the field coils. It is seen that for a

    4-pole motor, three speeds can be obtained easily.

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    2. Variable Resistance in Series with Motor: By increasing the resistance

    in series with the armature the voltage applied across the armature

    terminals can be decreased. With reduced voltage across the armature, the

    speed is reduced. However, it will be noted that since full motor current

    passes through this resistance, there is a considerable loss of power in it.

    Problems 41- A 230 V dc shunt motor runs at 800 rpm and takes armature current of 50 A.

    Find resistance to be added to the field circuit to increase speed to 1000 rpm

    at an armature current of 80 A. Assume flux proportional to field current.

    Armature resistance = 0.15 and field winding resistance = 250 .

    2- A 250 V, dc shunt motor has an armature resistance of 0.5 and a field

    resistance of 250 . When driving a load of constant torque at 600 rpm, the

    armature current is 20 A. If it is desired to raise the speed from 600 to 800

    rpm, what resistance should be inserted in the shunt field circuit? Assume that

    the magnetic circuit is unsaturated.

    3- A dc shunt motor takes an armature current of 20 A from a 220 V supply.

    Armature circuit resistance is 0.5 ohm. For reducing the speed by 50%,

    calculate the resistance required in the series, with the armature, if (a) the

    load torque is constant (b) the load torque is proportional to the square of the


    4- A 7.48 kW, 220 V, 990 rpm shunt motor has a full load efficiency of 88%, the

    armature resistance is 0.08 ohm and shunt field current is 2 A. If the speed of

    this motor is reduced to 450 rpm by inserting a resistance in the armature

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    circuit, find the motor output, the armature current, external resistance to be

    inserted in the armature circuit and overall efficiency. Assume the load torque

    to remain constant.

    5- A 250 V dc shunt motor has armature circuit resistance of 0.5 and a field

    circuit resistance of 125 . It drives a load at 1000 rpm and takes 30 A. The

    field circuit resistance is then slowly increased to 150 . If the flux and field

    current can be assumed to be proportional and if the load torque remains

    constant, calculate the final speed and armature current. [1186 rpm. 33.6 A]

    6- A shunt-wound motor has a field resistance of 400 and an armature

    resistance of 0.1 and runs off 240 V supply. The armature current is 60 A

    and the motor speed is 900 rpm; Assuming a straight line magnetization

    curve, calculate (a) the additional resistance in the field to increase the speed

    to 1000 rpm for the same armature current and (b) the speed with the original

    field current of 200 A. [(a) 44.4 (b) 842.5 rpm]

    7- A 250-V shunt motor has an armature current of 20 A when running at 1000

    rpm against full load torque. The armature resistance is 0.5 . What

    resistance must be inserted in series with the armature to reduce the speed to

    500 rpm at the same torque and what will be the speed if the load torque is

    halved with this resistance in the circuit? Assume the flux to remain constant

    throughout and neglect brush contact drop.

    8- A 7.46 kW, 220 V, 900 rpm shunt motor has a full-load efficiency of 88 per

    cent, an armature resistance of 0.08 and shunt field current of 2 A. If the

    speed of this motor is reduced to 450 rpm by inserting a resistance in the

    armature circuit, the load torque remaining constant, find the motor output,

    the armature current, the external resistance and the overall efficiency.

    9- A 200 V, dc series motor takes 40 A when running at 700 rpm. Calculate the

    speed at which the motor will run and the current taken from the supply if the

    field is shunted by a resistance equal to the field resistance and the load

    torque is increased by 50%. Armature resistance = 0.15 , field resistance =

    0.1 It may be assumed that flux per pole is proportional to the field.

    10- A 4-pole, 250 V dc series motor takes 20 A and runs at 900 rpm each field

    coil has resistance of 0.025 ohm and the resistance of the armature is 0.1 ohm.

    At what speed will the motor run developing the same torque if : (i) a diverter

    of 0.2 ohm is connected in parallel with the series field (ii) rearranging the

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    field coils in two series and parallel groups Assume unsaturated magnetic


    11- A 200V, dc series motor runs at 500rpm when taking a line current of 25A.

    The resistance of the armature is 0.2 and that of the series field 0.6. At

    what speed will it run when developing the same torque when armature

    diverter of 10 is used? Assume a straight line magnetization curve.[314rpm]

    Electric Braking: A motor and its load may be brought to rest quickly by

    using either (i) Friction Braking or (ii) Electric Braking. Mechanical

    brake has one drawback: it is difficult to achieve a smooth stop because

    it depends on the condition of the braking surface as well as on the skill

    of the operator. The excellent electric braking methods are available

    which eliminate